last two digits...
hey everyone, help me in solving this one.
Q1) What is last two digits of 8^7^6^5^4^3^2^1
thanks...
For easy follow up all the questions asked in different comments of this post are written below and later on these questions ‘ll be moved to Question Bank.
Q2) What are the last two digits of 122 x 123 x 125 x 127 x 129? Ans = 50
Q3)What are the last two digits of 15 x 37 x 63 x 51 x 97 x 17? Ans = 35
Q4) What will be the LAST THREE DIGITS in above qns?
What are the last two digits of
5) 3 to the 999th power i.e 3999
6) 14^14^14 i. e 14 14 14
7) 21 to the 100th power i.e 21100
8) what is the remainder when 32^33^34(i.e 323334) is divided by 7
what will be the last 2( and 3) digits of the following numbers
9) 411^27 (i.e 411 27)
10)109^21 (i.e 109 21)
11)465^102 (i.e 465 102)
12)what is the remainder when 22^33+10^35 (i.e 2233 + 1035 is divided by 45?
13)What unit digit of 2^3^4^5....
by Chandan - post : http://www.cat4mba.com/node/4922
Q1) What are the last two digits of 122 x 123 x 125 x 127 x 129?
solution: ny multiple of 25 should end in 25or50or75or00
furthr an odd multiple of 25 n 2 ends in 50
n even multiple of 25*2 ends in 00
hence 50
n/a
Hi,
Here is my work-put....Plz post the options so tht we can determine the ans from t.
6^anyting ends with 6
so,
the equ boils down to 8^7^6
7^6 ends with 9
8^9 ends with 8.
All the numbers/8 must be divisible by 4.so the last 2 digits must be divisible by 4.
so it shud be any1 of the below:
.......08
.........28
...........48
............68
.............88
So plz post the options..
-Deepak.
Hey, Your solution is wrong. The problem simplifies as 8^7^(a no. ending in 6). You've wrongly assumed that it is 7^6 and thus is ending in 9. It might be 7^(.....16) or 7^(......26) etc . Thus it might not end in 9. Thus ur solution is wrong...
I am not sure if the following is 100% correct and would love to see any flaw in the method.
Any how I would never attempt this sort of questions in real CAT.
32 = 9
49 = Something
5Something = x x x x. . . .25
Next target 6 x x x x. . . .25
61 = 6
62 = 36
63 = X16
64 =XX96
65 = XXX76
66 = XXXXX56
67 = XXXXXX36
Last two digits periodicity = 5
Now x x x . . . . 25 is divisible by 5.
Thus 6 x x x x. . . .25 = XXXXX. . . . .76
Next target 7XXXXX. . . . .76
71 = 7
72 = 49
73 = X43
74 = XX01
75 = XX07
Last two digits periodicity for 7 = 4
and 76 is divisible by 4
7XXXXX. . . . .76 = XXXXXXXX. . . . .01
Final target 8XXXXX. . . . .01
Last two digits periodicity for 8 = 20
64 44 24 04 84 64
12 52 92 32 72
96 16 36 56 76
68 28 88 48 8
So 8XXXXX. . . . .01 ends with 08 and thats theanswer
n/a
Wonderfully done, man !!!....I was thinking along the same lines but did not carry on till the end....only one small thing I would like to point out that while finding the periodicity of the last 2 digits for 8, you could have done with only 3 columns of calculation,
64 44 24 ....
12 52 92 ....
96 16 36 ....
68 28 88 ....
since in each row the tens digit increases or decreases by the same amount in consecutive columns and comes back to the original value after a cycle of 5 columns (in all the rows). So we can conclude that the periodicity is 20.
But I totally agree with you that we shudn't attempt such questions in cat...
is the answer of the 4th question is 01
last two digits of 21^100 is 01 please correct me if m wrong
guys help me solving this question
what is the remainder when 32^33^34 is divided by 7
a-4 b-2 c-3 d-1
reminder of 323334divided by 7
= reminder of 43334divided by 7
= reminder of 43K divided by 7
4 divided by 7 has a cylicity of 3 and the reminders are 4, 2, 1
So the answer is 1
Q1) What are the last two digits of 122 x 123 x 125 x 127 x 129? Ans = 50
Q2)What are the last two digits of 15 x 37 x 63 x 51 x 97 x 17? Ans = 35
Q3) What will be the LAST THREE DIGITS in above qns?
plz help....